Given a smooth manifold U, we have a map $\wedge^2\Gamma(U,TU)\to \Gamma(U,TU)$ given by $X\wedge Y\mapsto [X,Y]$, where $TU$ denotes the tangent bundle. Is it possible to describe the map $\Gamma(U,T^*U)\to \Gamma(U,\wedge^2 T^*U)$ corresponding to this map.

Though Victor's answer sort of tells the moral/ideological story of it, the question is still wrong as stated. The commutator of vector fields is not a well-defined map from the global sections of the exterior square of the tangent bundle, since it is not linear over the functions on the manifold, but only over the constants.
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Leonid PositselskiDec 8 '10 at 0:19

Additionally, the spaces $\Gamma(U,TU)$ and $\Gamma(U,T^{*}U)$ aren't dual to each other over the constants.
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Victor ProtsakDec 8 '10 at 1:59

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Well, when a map between the global sections of two vector bundles is linear over the functions, it does induce a map of the global sections of the dual bundles in the opposite direction. In this case, the exterior differential is not linear over the functions, and the commutator does not even exist as a map of global sections (of the kind stated in the question).
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Leonid PositselskiDec 8 '10 at 2:35

3 Answers
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To expand on Leonid's comment, if $\omega$ is a 1-form and $X,Y$ are vector fields, then
$$ d\omega(X \wedge Y) = X \omega(Y) - Y \omega(X) - \omega([X,Y]). $$
If the first two terms were not there, then one could say, as in Victor's answer, that the exterior derivative is (minus) the transpose of the Lie bracket of vector fields. The fact that the first two terms are there is symptomatic of Leonid's observation that the Lie bracket is not really a tensorial map.

What particular book did you have in mind?
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drbobmeisterDec 8 '10 at 2:46

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This is morally correct but technically wrong, as pointed out in Jose's answer and the comments to the question (including yours). But, yes, I do agree at a moral level that the bracket and the differential are dual. One way to make this precise, as you probably know but I'll mention it for other readers, is the Koszul duality between Lie algebra/oids and differential graded commutative algebras. When applied to tangent bundle, this Koszul duality returns the de Rham complex.
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Theo Johnson-FreydDec 8 '10 at 6:41